Electromagnetic Induction Test

Result

Electromagnetic Induction Test - 7

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0.25

A flux of 1m Wb passes through a strip having an area A = 0.02 m² . The plane of the strip is at an angle of 60 ºto the direction of a uniform field B. The value of B is

A
0.1 T

B
0.058 T

C
4.0 mT

D
none of the above

Solution

formula of flux is = BAcos θ
θis angle between A and B
θ=30 ∘
ϕ=1 ×10⁻³ wb
A=0.02 m²
angle between screen and B is 60 ∘
B = ϕ/acos30 = 10^-3 / 2 x 10^-2 x cos √3/2 = 0.058T
Question 2
1 / -0.25

The number of turns in a long solenoid is 500. The area of cross-section of solenoid is 2 ×10^-3 m² . If the value of magnetic induction, on passing a current of 2 amp, through it is 5 ×10^-3 tesla, the magnitude of magnetic flux connected with it in webers will be

A
5 ×10^-3

B
10^-2

C
10^-5

D
2.5

Solution

Φ=BAN
=5x10^-3 x2x10^-3 x500
=5x10^-3 x1000x10^-3
=5x10^-3
Question 3
1 / -0.25

A conducting loop of radius R is present in a uniform magnetic field B perpendicular the plane of the ring. If radius R varies as a function of time `t ', as R = R₀ + t. The e.m.f induced in the loop is

A
2 π(R₀ + t) B clockwise

B
π(R₀ + t)B clockwise

C
2 π(R₀ + t)B anticlockwise

D
zero

Solution

ε=−d ϕ/dt
=−β(dA/dt)
⇒∣ε∣=βπd(R₀ +t)²
=2 πβ(R₀ +t)
Since inward flux is increasing the EMF will be induced counter clockwise.
Question 4
1 / -0.25

The instantaneous flux associated with a closed circuit of 10 Ωresistance is indicated by the following reaction f = 6t^{2 _} 5t + 1, then the value in amperes of the induced current at t = 0.25 sec will be

A
1.2

B
0.8

C
6

D
0.2

Solution

ϕ=6t² −5t+1
e= d ϕ/dt =12t −5
at t=0.25sec
∣e ∣=(12/4 −5)=∣3 −5 ∣=2
i=e/R=2/10=0.2A
Question 5
1 / -0.25

When a magnet is being moved towards a coil, the induced emf does not depend upon

A
the number of turns of the coil

B
the motion of the magnet

C
the magnetic moment of the magnet

D
the resistance of the coil

Solution

The magnitude of induced emf is independent of the resistance of the coil.
∴e=−d ϕ/dt=−N (d/dt) [(BA)/dt]
Thus, e depends upon N, B and A.
But induced current depends upon R.
Question 6
1 / -0.25

An electron is moving in a circular orbit of radius R with an angular acceleration a. At the centre of the orbit is kept a conducting loop of radius r, (r <

A
zero, since charge on electron in constant

B

C

D
none of these
Question 7
1 / -0.25

A field of 5 x 10⁴ /p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10^-2 m² . The coil is removed from the field in 0.1 second. Then the induced emf in the coil is

A
0.1 V

B
80 KV

C
7.96 V

D
none of the above

Solution
Question 8
1 / -0.25

square wire loop of 10.0 cm side lies at right angles to a uniform magnetic field of 20T. A 10 V light bulb is in a series with the loop as shown in the fig. The magnetic field is decreasing steadily to zero over a time interval Dt. The bulb will shine with full brightness if Dt is equal to

A
20 ms

B
0.02 ms

C
2 ms

D
0.2 ms

Solution

e=−Δϕ/Δt=ϕi −ϕf/Δt
δt=(ϕ_i −ϕ_f )/e=(BA −0)/e=(20 ×10⁻² )/10
=20 ×10⁻³ sec=20ms
Question 9
1 / -0.25

The rate of change of magnetic flux denisty through a circular coil of area 10^-2 m² and number of turns 100 is 10³ Wb/m² /s. The value of induced e.m.f. will be

A
10^-2 V

B
10^-3 V

C
10V

D
10³ V

Solution
Question 10
1 / -0.25

A long solenoid contains 1000 turns/cm and an alternating current of peak value 1A is flowing in it. A search coil of area of cross-section 1 ×10^-4 m² and having 50 turns is placed inside the solenoid with its plane perpendicular to the axis of the solenoid. A peak voltage of 2p² ×10^-2 V is produced in the search coil. The frequency of current in the solenoid will be

A
50 Hz

B
100 Hz

C
500 Hz

D
1000 Hz